Work, Energy, and Simple Machines Class 9 Notes, Numericals and Questions

What Work, Energy, and Simple Machines is about

Weightage: This chapter is part of Unit III, Motion, Force, Work and Sound, which carries 23 of the 80 theory marks in the school annual exam, shared with Describing Motion Around Us, How Forces Affect Motion and Sound Waves: Characteristics and Applications.Source: CBSE Science (086) syllabus, Class IX 2026-27, course structure, Unit III Motion, Force, Work and Sound

Key concepts

Work done by a constant force

NCERT section 7.1 Work Done by a Constant Force

Lifting one 5 kg bag through 1 m is some work; lifting three bags, or one bag through 3 m, is three times as much. So work grows with force and with distance moved along the force.

$W=F\times s$

where $s$ is the displacement in the direction of the force.

• SI unit: joule (J). 1 J is the work done when a 1 N force moves an object 1 m along the force. $1\text{ J}=1\text{ N m}=1{\text{ kg m}}^2{\text{ s}}^{-2}$.
• Always say which force (agency) does work on which object.
• On a force-displacement graph, work = area under the graph (a rectangle for a constant force; useful even when the force varies).
• Work has no direction, but it can be positive or negative.

Zero work when:

• there is no force ($F=0$);
• there is no displacement (pushing a wall, holding a barbell still: you feel tired because muscles use internal energy, but no work is done on the object);
• the force is perpendicular to the displacement (carrying a box while walking on level ground: the upward force does no work).

Worked example

(i) A girl lifts an 8 kg bag of rice slowly through 1.5 m. Find the work she does ($g=9.8$ m s${}^{-2}$). (ii) She then walks 20 m along a level corridor holding the bag. How much work does her upward force do now?

Try it first, then show the solutionSolution

1. (i) Force needed $=mg=8\times 9.8=78.4$ N upwards.
2. $W=F\times s=78.4\times 1.5=117.6$ J.
3. (ii) Her force is upwards, the displacement is horizontal: they are perpendicular.
4. So the work done by that force is zero.

Answer: (i) 117.6 J (ii) zero.

Positive and negative work

NCERT section 7.1 Work Done by a Constant Force

• Positive work: force and displacement in the same direction (pushing a wheelchair forward; lifting a dumbbell).
• Negative work: force opposite to the displacement (a goalkeeper stopping a ball; friction on a sliding object; lowering a dumbbell slowly while holding it; brakes on a car).

When you write $W=F\times s$, take the displacement as negative if it is opposite to the force.

Forces between two objects come in pairs (third law), so both do work: when a goalkeeper stops a ball, she does negative work on the ball while the ball does positive work on her hands.

Worked example

A car's brakes exert a steady 2000 N backward force while the car moves 15 m forward before stopping. Find the work done by the brakes and state its sign.

Try it first, then show the solutionSolution

1. The force is backward and the displacement forward, so displacement along the force $=-15$ m.
2. $W=2000\times (-15)=-30000$ J.
3. Negative: the brakes take energy away from the car.

Answer: −30 000 J (negative work).

Work-energy theorem and forms of energy

NCERT section 7.2 The Work-Energy Theorem

An object that can do work has energy. A thrown ball can knock down the wickets; a raised flowerpot can break something below if it falls. They got this energy because work was done on them.

Work-energy theorem: work done on an object = change in its energy.

• Positive work → energy increases; negative work → energy decreases.
• It also holds for a system of objects and for forces that are not constant.
• SI unit of energy: joule, named after James Prescott Joule, who showed how mechanical and thermal energy are related.

Work is one way to transfer energy; energy also moves as heat, by radiation (Sun to Earth), in electric circuits and in sound waves.

Forms of energy: mechanical (motion or position), thermal, light, sound, electrical, chemical (stored in bonds in food and fuel), nuclear (in atomic nuclei).

Conversions: bulb: electrical → light (and heat); heater: electrical → thermal; muscles: chemical → mechanical; bell: mechanical → sound; solar panel: light → electrical; photosynthesis: light → chemical; burning matchstick: chemical → heat and light.

In a carrom shot, the striker does positive work on the white coin (its energy rises) and the coin does negative work on the striker (its energy falls).

Worked example

A 2 kg ball rolling at 10 m s${}^{-1}$ on grass comes to rest after 5 m. Using the work-energy theorem, find the average friction force.

Try it first, then show the solutionSolution

1. Initial kinetic energy $=\frac{1}{2}\times 2\times 10^2=100$ J; final $=0$.
2. Change in energy $=-100$ J = work done by friction.
3. $-100=F\times (-5)$, so $F=20$ N, opposite to the motion.

Answer: 20 N.

Kinetic energy

NCERT section 7.4 Mechanical Energy

Kinetic energy (K) is energy due to motion. An object at rest has zero kinetic energy.

Derivation: a constant force $F$ acts on mass $m$, changing its velocity from $u$ to $v$ over displacement $s$.
1. $v^2=u^2+2as$, so $s=\frac{v^2-u^2}{2a}$.
2. $W=F\times s=ma\times \frac{v^2-u^2}{2a}=\frac{1}{2}m(v^2-u^2)$.
3. By the work-energy theorem this is the change in energy. Starting from rest ($u=0$):

$K=\frac{1}{2}m{v}^2$

• Unit: joule; no direction.
• $K$ depends on $v^2$: doubling the speed makes the kinetic energy four times; doubling the mass doubles it.
• If no work is done, kinetic energy stays the same (constant velocity).

A cricket ball of about 0.2 kg bowled at 154.8 km h${}^{-1}$ (43 m s${}^{-1}$) has about 185 J of kinetic energy.

Worked example

Find the kinetic energy of (i) a 60 kg runner at 8 m s${}^{-1}$ (ii) a 150 kg scooter with rider at 54 km h${}^{-1}$. What happens to the scooter's kinetic energy if its speed halves?

Try it first, then show the solutionSolution

1. (i) $K=\frac{1}{2}\times 60\times 8^2=1920$ J.
2. (ii) $54$ km h${}^{-1}$ $=15$ m s${}^{-1}$; $K=\frac{1}{2}\times 150\times 15^2=16875$ J.
3. Halving $v$ makes $v^2$ one-quarter, so $K$ becomes $\frac{16875}{4}\approx 4219$ J.

Answer: (i) 1920 J (ii) 16 875 J; it falls to one-quarter.

Potential energy

NCERT section 7.4 Mechanical Energy

Potential energy is energy stored in an object because of its deformed shape, or in a system because of the relative positions of its parts.

• Elastic: a stretched slingshot (*gulel*) band, a bent bow, a compressed spring. Work done to deform them is stored and released as kinetic energy of the stone, arrow or object.
• Magnetic and electric: pulling apart unlike magnetic poles or opposite charges stores energy; when released they rush together.
• Gravitational: lifting a ball away from the Earth stores energy in the Earth-ball system (we simply call it the ball's potential energy because the Earth hardly moves).

Activity 7.1: a heavy ball dropped into sand from 2 m makes a deeper pit than from 1 m: more height, more stored energy.

Formula: to lift mass $m$ slowly through height $h$, the force is $mg$, so $W=mgh$, and this is stored:

$U=mgh$

(taking $U=0$ at the ground; valid near the Earth's surface, where $g$ is nearly constant).

Potential energy depends only on the height, not on the path taken (lift or stairs give the same gain). Work done against friction is not stored as potential energy.

Worked example

A 40 kg student climbs to the top of a 15 m building, once by the stairs and once in a lift. Find her gain in potential energy each time ($g=10$ m s${}^{-2}$). What does this show?

Try it first, then show the solutionSolution

1. $U=mgh=40\times 10\times 15=6000$ J.
2. Both paths end at the same height, so both give 6000 J.
3. Gravitational potential energy depends only on height, not on the path.

Answer: 6000 J both ways; potential energy is independent of the path.

Conservation of mechanical energy

NCERT section 7.4 Mechanical Energy

Mechanical energy = kinetic energy + potential energy.

For an object falling freely from height $h$ (starting from rest):

• At the top: $U=mgh$, $K=0$, total $=mgh$.
• After time $t$: $v=gt$ and height $h^{\prime }=h-\frac{1}{2}g{t}^2$, so $U=mgh-\frac{1}{2}m{g}^2{t}^2$ and $K=\frac{1}{2}m{g}^2{t}^2$. Total $=mgh$ again.
• Just before landing: $U=0$, $K=mgh$.

Potential energy lost = kinetic energy gained. If only gravity does work, mechanical energy stays constant. This is the conservation of mechanical energy.

Pendulum (Activity 7.2): at the extreme point P: only potential energy; at the lowest point Q: only kinetic energy (fastest); at R on the other side: only potential energy again, almost at the same height. It slowly stops because of friction at the support and air resistance.

Slide: $\frac{1}{2}m{v}^2=mgh$ gives $v=\sqrt{2gh}$, which does not depend on the child's mass or the shape of the slide (ignoring friction).

The total energy of an isolated system is always conserved; mechanical energy may change into heat and sound through friction. A perpetual motion machine is impossible.

In the Himalayas, a *gharat* or *panchakki* uses falling water (potential → kinetic energy) to turn a grinding stone; dams use the same idea for electricity.

Worked example

A stone is dropped from 20 m ($g=10$ m s${}^{-2}$). Use energy to find its speed (i) when it is 5 m above the ground (ii) just before it lands.

Try it first, then show the solutionSolution

1. Per kg: total energy $=gh=10\times 20=200$ J kg${}^{-1}$.
2. (i) It has fallen 15 m: $\frac{1}{2}{v}^2=10\times 15=150$, so $v=\sqrt{300}\approx 17.3$ m s${}^{-1}$.
3. (ii) It has fallen 20 m: $\frac{1}{2}{v}^2=200$, so $v=\sqrt{400}=20$ m s${}^{-1}$.

Answer: (i) about 17.3 m s${}^{-1}$ (ii) 20 m s${}^{-1}$; the mass does not matter.

Power

NCERT section 7.5 Power

Running up the stairs in one minute and walking up in five minutes need the same work, but running needs more power.

Power is the rate of doing work:

$P=\frac{W}{t}$

• SI unit: watt (W); $1\text{ W}=1{\text{ J s}}^{-1}$. Named after James Watt, who built an efficient steam engine.
• 1 horsepower (hp) = 746 W, still used for engines and pumps.
• More work in the same time, or the same work in less time, means more power.
• When a machine gives an object speed, the work done equals the gain in kinetic energy, so $P=\frac{\frac{1}{2}m{v}^2}{t}$ (ignoring losses).

Worked example

A pump lifts 500 kg of water to a tank 12 m high in 60 s ($g=10$ m s${}^{-2}$). Find its power in watts and in horsepower.

Try it first, then show the solutionSolution

1. Work $=mgh=500\times 10\times 12=60000$ J.
2. $P=\frac{60000}{60}=1000$ W.
3. In hp: $\frac{1000}{746}\approx 1.34$ hp.

Answer: 1000 W, about 1.34 hp.

Simple machines: pulley and inclined plane

NCERT section 7.6 Simple Machines

A simple machine makes a task easier by changing the size or direction of the force needed. It cannot reduce the total work (ignoring friction, work put in = useful work out).

• Effort: the force we apply. Load: the force to be overcome.
• Mechanical advantage (MA) $=\frac{\text{load}}{\text{effort}}$ (no unit).

Pulley: a grooved wheel guiding a rope.

• A fixed pulley (flagpole, well) only changes the direction of the force: pulling down is easier than lifting up. Effort = load, so MA = 1.
• Movable pulleys and pulley systems (cranes, lifts) give MA greater than 1.

Inclined plane (ramp): pushing a load of weight $mg$ up a smooth ramp of length $L$ to height $h$:

work in = work out: $F^{\prime }\times L=mgh$, so $\text{MA}=\frac{mg}{F^{\prime }}=\frac{L}{h}$

Since $L>h$, MA > 1. A longer, gentler slope needs less force but over a greater distance (Activity 7.3 with a spring balance and a toy car). That is why hill roads wind round in gentle slopes and why a sloping ladder is easier than a vertical one.

Worked example

A 3 m smooth plank is used to load a 600 N drum onto a truck bed 0.75 m high. Find the mechanical advantage and the effort. How much work is done, by ramp and by direct lifting?

Try it first, then show the solutionSolution

1. MA $=\frac{L}{h}=\frac{3}{0.75}=4$.
2. Effort $=\frac{\text{load}}{\text{MA}}=\frac{600}{4}=150$ N.
3. Work by ramp $=150\times 3=450$ J; by lifting $=600\times 0.75=450$ J: the same.

Answer: MA 4; effort 150 N; 450 J either way (the ramp reduces force, not work).

Levers

NCERT section 7.6 Simple Machines

A lever is a rigid bar that turns about a fixed point, the fulcrum.

• Effort arm: distance of the effort from the fulcrum. Load arm: distance of the load from the fulcrum.
• The end with the small effort moves a large distance; the end with the load moves a small distance. Work in = work out: $F_1{d}_1=F_2{d}_2$.

Law of the lever (Activity 7.5 with a beam balance of coins):

effort × effort arm = load × load arm

$\text{MA}=\frac{\text{load}}{\text{effort}}=\frac{\text{effort arm}}{\text{load arm}}$

A longer effort arm means less effort.

Classes of levers:

Class	In the middle	Examples	MA
I	fulcrum	seesaw, crowbar, scissors, pliers, beam balance	more than, equal to or less than 1
II	load	bottle opener, wheelbarrow, lemon squeezer	always more than 1
III	effort	tongs, tweezers, broom, fishing rod, oar, forearm	always less than 1 (gains speed or convenience)

A spoon prising open a tin lid, and cutting hard things close to the scissors' pivot, both use a short load arm to get a large force. Many machines combine several simple machines.

Worked example

A crowbar 1.4 m long has its fulcrum 0.2 m from the rock end. The rock exerts a load of 900 N. Find the mechanical advantage and the effort needed at the other end. Which class of lever is this?

Try it first, then show the solutionSolution

1. Load arm $=0.2$ m; effort arm $=1.4-0.2=1.2$ m.
2. MA $=\frac{1.2}{0.2}=6$.
3. Effort $=\frac{900}{6}=150$ N.
4. Fulcrum between load and effort: Class I.

Answer: MA 6; effort 150 N; Class I lever.

Science in real life: Work, Energy, and Simple Machines

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Power, Pulleys, Inclined Planes and Levers
• Work, Kinetic Energy, Potential Energy and Conservation

Work, Energy, and Simple Machines: formulas and key facts

Work

$W=F\times s$ (s along the force); 1 J = 1 N × 1 m

Work-energy theorem

Work done on an object = change in its energy; $W=\frac{1}{2}m(v^2-u^2)$

Kinetic energy

$K=\frac{1}{2}m{v}^2$

Gravitational potential energy

$U=mgh$ (near the Earth's surface)

Conservation of mechanical energy

$K+U$ = constant when only gravity does work; speed after falling $h$: $v=\sqrt{2gh}$

Power

$P=\frac{W}{t}$; 1 W = 1 J s⁻¹; 1 hp = 746 W

Mechanical advantage

$\text{MA}=\frac{\text{load}}{\text{effort}}$

Inclined plane

$\text{MA}=\frac{L}{h}$ (smooth ramp)

Lever

Effort × effort arm = load × load arm; $\text{MA}=\frac{\text{effort arm}}{\text{load arm}}$

Fixed pulley

$MA=1(changesdirectiononly).$

Zero work

No force, no displacement, or force perpendicular to displacement.

Work, Energy, and Simple Machines: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Work, Energy, and Simple Machines

The question patterns to expect, and the method that works for each.

Tricks to remember Work, Energy, and Simple Machines

Short memory aids. Say each one aloud once, then use it on a question the same day.

• No move, no work

  If it does not move, or moves sideways to the force, the force does no work.

  Use it when: Zero-work questions.

• Half m v squared

  Double speed, four times the energy; double mass, double the energy.

  Use it when: KE comparison questions.

• Top is potential, bottom is kinetic

  Pendulums, slides and falling stones swap PE for KE on the way down and back on the way up.

  Use it when: Conservation of energy.

• FLE 1-2-3

  In the middle: Fulcrum (class I), Load (class II), Effort (class III).

  Use it when: Classifying levers.

• Machines trade, not make

  A machine trades a smaller force for a longer distance; work in = work out.

  Use it when: Reasoning on why machines do not save energy.

Work, Energy, and Simple Machines: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Work, Energy, and Simple Machines

The slips examiners see most often, and what to write instead.

• ✗Mistake: Saying work is done when holding a load still

  Holding a load makes us tired.

  Wrong working

  A man holding a 20 kg suitcase still for 5 minutes does work $=20\times 9.8\times 300$.

  Right working

  Displacement is zero, so work on the suitcase is zero; tiredness is from internal energy used by muscles.

  ✓Do this: Work needs displacement along the force.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting to square the speed or to halve

  Rushing the formula.

  Wrong working

  $K=\frac{1}{2}\times 2\times 10=10$ J for a 2 kg ball at 10 m s${}^{-1}$.

  Right working

  $K=\frac{1}{2}\times 2\times 10^2=100$ J.

  ✓Do this: Write the formula in full before substituting.

  Usually costs: The answer mark.

• ✗Mistake: Using mass instead of weight in $W=Fs$ or $U=mgh$

  Mass and weight are confused.

  Wrong working

  Lifting a 5 kg bag by 2 m: $W=5\times 2=10$ J.

  Right working

  $W=mgh=5\times 9.8\times 2=98$ J.

  ✓Do this: Force in newtons: multiply mass by $g$.

  Usually costs: The answer mark.

• ✗Mistake: Claiming a machine reduces work

  Machines make tasks feel easier.

  Wrong working

  A ramp reduces the work needed to load a truck.

  Right working

  A ramp reduces the force but increases the distance; the work (ignoring friction) is the same, and friction makes it slightly more.

  ✓Do this: Machines reduce force or change direction, not work.

  Usually costs: 1 to 2 marks in reasoning questions.

• ✗Mistake: Getting the lever arms the wrong way round

  Students divide load arm by effort arm.

  Wrong working

  Effort arm 1.2 m, load arm 0.2 m: MA $=\frac{0.2}{1.2}$.

  Right working

  MA $=\frac{\text{effort arm}}{\text{load arm}}=\frac{1.2}{0.2}=6$.

  ✓Do this: A good lever has MA greater than 1 when the effort arm is longer.

  Usually costs: The method and answer marks.

NCERT exercise guide: Work, Energy, and Simple Machines

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Work, Energy, and Simple Machines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Work done by a force is zero when the angle between force and displacement is
   (a) 0°
   (b) 90°
   (c) 180°
   (d) 45°

   Hint

   Carrying a bag on level ground.

   Answer

   (b) 90°

2. Q2 · 1 mark · MCQExam level

   A wheelbarrow is a lever of
   (a) class I
   (b) class II
   (c) class III
   (d) no class

   Hint

   Load in the middle.

   Answer

   (b) class II

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): Doubling the speed of a car doubles its kinetic energy.
   Reason (R): Kinetic energy is proportional to the square of the speed.
   (a) A and R are both true, and R explains A correctly.
   (b) A and R are both true, but R does not explain A.
   (c) A is true, R is false.
   (d) A is false, R is true.

   Hint

   $v^2$.

   Answer

   (d). Doubling speed makes KE four times, so A is false; R is true.

4. Q4 · 2 marks · Very short answerExam level

   Name the energy change in: (i) an electric bulb (ii) a solar panel (iii) photosynthesis (iv) a burning matchstick.

   Hint

   From → to.

   Answer

   (i) Electrical → light and heat (ii) light → electrical (iii) light → chemical (iv) chemical → heat and light.

5. Q5 · 2 marks · Very short answerHarder exam level

   A trolley is pushed 12 m along a level floor with a steady 40 N force. How much work is done by the push? If the trolley moves at constant speed, how much work does friction do?

   Hint

   Constant speed means no change in KE.

   Answer

   Push: $40\times 12=480$ J. Constant speed: KE unchanged, so friction does $-480$ J.

6. Q6 · 3 marks · Short answerHarder exam level

   On Mars, $g$ is about 3.7 m s⁻². A ball is thrown straight up at 7.4 m s⁻¹ on Mars and at the same speed on Earth ($g=9.8$ m s⁻²). Use energy to find how high it rises on each, ignoring air.

   Hint

   $\frac{1}{2}{v}^2=gh$.

   Answer

   Mars: $h=\frac{v^2}{2g}=\frac{54.76}{7.4}=7.4$ m. Earth: $h=\frac{54.76}{19.6}\approx 2.8$ m. The same kinetic energy lifts it higher where $g$ is smaller.

7. Q7 · 3 marks · Short answerHarder exam level

   A 75 kg weightlifter lifts a 75 kg barbell through 2 m in 5 s ($g=10$ m s⁻²). Find the work done on the barbell and the power. Is any work done on the barbell while she holds it above her head?

   Hint

   $mgh$ then divide by time.

   Answer

   $W=75\times 10\times 2=1500$ J; $P=\frac{1500}{5}=300$ W. Holding it still: no displacement, so no work on the barbell.

8. Q8 · 3 marks · Short answerHarder exam level

   A smooth ramp 2.4 m long is used to wheel a 480 N load into a van whose floor is 0.6 m high. Find the mechanical advantage and the effort needed. Why are wheelchair ramps built long and gentle?

   Hint

   $\text{MA}=\frac{L}{h}$.

   Answer

   MA $=\frac{2.4}{0.6}=4$; effort $=\frac{480}{4}=120$ N. A long, gentle ramp has a large $\frac{L}{h}$, so a wheelchair user needs only a small force, though over a longer distance.

9. Q9 · 3 marks · Short answerHarder exam level

   Why do you cut a hard twig near the pivot of a pair of scissors? Why is it easier to open a tin lid with a spoon than with your fingers?

   Hint

   Short load arm.

   Answer

   Near the pivot the load arm is short and the effort arm (handle) is long, so MA is large and the scissors apply a large force. A spoon used as a lever has a long effort arm and the lid edge as a short load arm, multiplying your effort.

10. Q10 · 3 marks · Short answerHarder exam level

    Describe the energy changes of a ball on a science-park roller track as it goes from the highest point A down to B and up to C. Why is each later hump lower than the one before?

    Hint

    Friction.

    Answer

    At A: mostly PE. Going down to B: PE changes to KE (fastest at the bottom). Going up to C: KE changes back to PE. Friction and air resistance take a little energy at each stage as heat and sound, so later humps must be lower for the ball to get over them.

11. Q11 · 3 marks · Short answerHOTS / competency

    A 5 kg box slides on a smooth floor at 4 m s⁻¹. A 15 N force then acts against its motion over 2 m. Find its speed after the 2 m, and how much further it would go if the force kept acting.

    Hint

    Work-energy theorem.

    Answer

    Initial KE $=\frac{1}{2}\times 5\times 16=40$ J. Work by the force $=-15\times 2=-30$ J, so KE $=10$ J and $v=\sqrt{\frac{2\times 10}{5}}=2$ m s⁻¹. To remove the last 10 J: $\frac{10}{15}\approx 0.67$ m further.

12. Q12 · 3 marks · Short answerHOTS / competency

    A 45 kg student walks down a 9 m staircase at a steady speed ($g=10$ m s⁻²). Find the work done on her by gravity. Her kinetic energy does not increase: explain, using the work-energy theorem, where this energy goes.

    Hint

    Total work = change in KE = 0.

    Answer

    Gravity does $+45\times 10\times 9=+4050$ J. Since her KE does not change, the total work on her is zero, so her leg muscles do $-4050$ J by braking each step. This energy ends up mostly as heat in her muscles and the stairs (and a little sound).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define one joule.Show answer

   The work done when a force of 1 N moves an object 1 m in the direction of the force.

2. 2.Give two situations where a force does no work.Show answer

   Pushing a rigid wall (no displacement); carrying a bag on level ground (force perpendicular to displacement).

3. 3.State the work-energy theorem.Show answer

   The work done on an object equals the change in its energy.

4. 4.Write the formula for kinetic energy.Show answer

   $K=\frac{1}{2}m{v}^2$.

5. 5.What is potential energy?Show answer

   Energy stored in an object due to its deformation, or in a system due to the relative positions of its parts.

6. 6.What is mechanical energy?Show answer

   The sum of kinetic and potential energy.

7. 7.What is the SI unit of power, and how many watts make one horsepower?Show answer

   The watt; 1 hp = 746 W.

8. 8.What is mechanical advantage?Show answer

   Load ÷ effort.

9. 9.What is the MA of a fixed pulley?Show answer

   1; it only changes the direction of the force.

10. 10.Give one example each of class I, II and III levers.Show answer

    Seesaw (I), wheelbarrow (II), tweezers (III).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you calculate work for lifting and pushing, with the right sign?
• ☐Can you explain three cases of zero work?
• ☐Can you state and use the work-energy theorem?
• ☐Can you derive $K=\frac{1}{2}m{v}^2$?
• ☐Can you calculate PE and show it is path-independent?
• ☐Can you show mechanical energy is constant for a falling body?
• ☐Can you describe energy changes in a pendulum?
• ☐Can you calculate power in watts and horsepower?
• ☐Can you find MA for a ramp and a lever?
• ☐Can you classify levers and give examples?

Work, Energy, and Simple Machines: frequently asked questions